The system
\begin{align} \dfrac{dx}{dt}&=-8 - 4 y + 2 x y, \\ \dfrac{dy}{dt}&=-x^2 + 4 y^2. \end{align}
I have linearized it for one of the singularity point(-1,-2):
\begin{align} \dfrac{dx}{dt}&=-2x+4, \\ \dfrac{dy}{dt}&=4x-8y. \end{align}
The problem is that I got a free member(don't know how it in English) - "4" in the first equation(i.e. $-2x+0y+4$).
user247327 noticed that my linearization was wrong, so I provide a linearization respectively to the second singularity point(4,2)
\begin{align} \dfrac{dx}{dt}&=4x+4y-24, \\ \dfrac{dy}{dt}&=-8x+16y+60. \end{align}
Equations have -24 and 60 free members. The equation for lamdas(eigenvalues), I've found is
\begin{align} \lambda_{1,2}=\dfrac{a+d}{2}+-\sqrt{\dfrac{(a+d)^2-4(ad-bc)}{4}} \end{align}
where a,b,c,d are the coefficient of the linear equations.
Please, explain what should I do in such cases? The goal I want to reach is to determine type of the singularity points of nonlinear system.
Your "$\frac{dx}{dt}= -2x+ 4$"should be $\frac{dx}{dt}= -2x+ 4y$. And if $\lambda$ is an eigenvalue, we must have $\frac{dx}{dt}= -2x+ 4y= \lambda x$ and $\frac{dy}{dt}= 4x- 8y= \lambda y$. The two equations, $-2x+ 4y= \lambda x$ and $4x- 8y= \lambda y$ are equivalent to $(-2- \lambda)x+ 4y= 0$ and $4x- (8+ \lambda)y= 0$. Those equations have the obvious trivial solution $x= y= 0$. An "eigenvalue" for this problem is a value of $\lambda$ such that this equation has other, non-trivial solutions.
This set of equations can also be written as the matrix equation, $$\begin{bmatrix}-2- \lambda & 4 \\ 4 & -8- \lambda \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ IF that matrix is invertible we could easily solve it by multiplying both sides by that inverse to get the trivial solution $$\begin{bmatrix}x \\ y \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ There will be a non-trivial solution only if that matrix is NOT invertible which means that its determinant must be 0.
That is, $\lambda$ is an eigenvalue only if $$\left|\begin{array}{cc}-2- \lambda & 4 \\ 4 & -8- \lambda \end{array}\right|= (-2- \lambda)(-8- \lambda)- 16= \lambda^2+ 10\lambda= \lambda(\lambda+ 10)= 0.$$ That is, the eigenvalues are 0 and -10.